CS 268: Lectures 13/14 (Route Lookup and Packet Classification)

CS 268: Lectures 13/14(Route Lookup and Packet Classification) Ion Stoica April 1/3, 2002

Lookup Problem • Identify the output interface to forward an incoming packet based on its destination address • Routing (forwarding) tables summarize information by maintaining prefixes • Route lookup  find the longest prefix in the table that matches the packet destination address istoica@cs.berkeley.edu

12.82.100.101 128.16.120.111 Example • Packet with destination address 12.82.100.101 is sent to interface 2, as 12.82.100.xxx is the longest prefix matching packet’s destination address 128.16.120.xxx 1 12.82.xxx.xxx 3 12.82.100.xxx 2 … … 1 2 istoica@cs.berkeley.edu

Patricia Tries • Use binary tree paths to encode prefixes • Advantage: simple to implement • Disadvantage: one lookup may take O(m), where m is number of bits (32 in the case of IPv4) 1 0 001xx 2 0100x 3 10xxx 1 01100 5 1 0 0 1 0 1 1 2 0 0 3 0 5 istoica@cs.berkeley.edu

Lulea’s Routing Lookup Algorithm • Minimize number of memory accesses • Minimize size of data structure • Small size allow to fit entire data structure in the cache (why do you care about size?) • Solution: use a three level data structure istoica@cs.berkeley.edu

root heads genuine heads First Level: Bit-Vector • Cover all prefixes down to depth 16 • Use one bit to encode each prefix • Memory requirements: 216 = 64 Kb = 8 KB istoica@cs.berkeley.edu

First Level: Pointers • Maintain 16-bit pointers to (1) next-hop (routing) table or (2) to two level chuncks • 2 bits encode pointer type • 14 bits represent an index into routing table or into an array containing level two chuncks • Pointers are stored at consecutive memory addresses • Problem: find the pointer istoica@cs.berkeley.edu

Problem: find pointer Example 0006abcd 000acdef bit vector … 1 0 0 0 1 0 1 1 1 0 0 0 1 1 1 1 pointer array … Routing table Level two chunks istoica@cs.berkeley.edu

Code Word and Base Indexes Array • Split the bit-vector in bit-masks (16 bits each) • Find corresponding bin-mask • How? • Maintain a16-bit code word for each bit-mask (10-bit value; 6-bit offset) • Maintain a base index array (one 16-bit entry for each 4 code words) number of previous ones in the bit-vector Bit-vector Code word array Base index array istoica@cs.berkeley.edu

First Level: Finding Pointer Group • Use first 12 bits to index into code word array • Use first 10 bits to index into base index array first 12 bits 4 address: 004C 1 first 10 bits Code word array Base index array = 13 + istoica@cs.berkeley.edu

First Level: Encoding Bit-masks • Observation: not all 16-bit values are possible • Example: bit-mask 1001… is not possible (why?) • Let a(n) be number of bit-masks of length 2n • Compute a(n) using recurrence: • a(0) = 1 • a(n) = 1 + a(n-1)2 • For length 16, we get only 677 possible values for bit-masks • This can be encoded in 10 bits • Values ri in code words • Store all possible bit-masks in a table, called maptable istoica@cs.berkeley.edu

First Level: Finding Pointer Index • Each entry in Maptable is an offset of 4 bits: • Offset of pointer in the group • Number of memory accesses: 3 (7 bytes accessed) istoica@cs.berkeley.edu

First Level: Memory Requirements • Code word array: one code word per bit-mask • 64 Kb • Based index array: one base index per four bit-mask • 16 Kb • Maptable: 677x16 entries, 4 bits each • ~ 43.3 Kb • Total: 123.3 Kb = 15.4 KB istoica@cs.berkeley.edu

First Level: Optimizations • Reduce number of entries in Maptable by two: • Don’t store bit-masks 0 and 1; instead encode pointers directly into code word • If r value in code word larger than 676  direct encoding • For direct encoding use r value + 6-bit offset istoica@cs.berkeley.edu

Levels 2 and 3 • Levels 2 and 3 consists of chunks • A chunck covers a sub-tree of height 8  at most 256 heads • Three types of chunks • Sparse: 1-8 heads • 8-bit indices, eight pointers (24 B) • Dense: 9-64 heads • Like level 1, but only one base index (< 162 B) • Very dense: 65-256 heads • Like level 1 (< 552 B) • Only 7 bytes are accessed to search each of levels 2 and 3 istoica@cs.berkeley.edu

Limitations • Only 214 chuncks of each kind • Can accommodate a growth factor of 16 • Only 16-bit base indices • Can accommodate a growth factor of 3-5 • Number of next hops <= 214 istoica@cs.berkeley.edu

Notes • This data structure trades the table construction time for lookup time (build time < 100 ms) • Good trade-off because routes are not supposed to change often • Lookup performance: • Worst-case: 101 cycles • A 200 MHz Pentium Pro can do at least 2 millions lookups per second • On average: ~ 50 cycles • Open question: how effective is this data structure in the case of IPv6 ? istoica@cs.berkeley.edu

Classification Problem • Classify an IP packet based on a number of fields in the packet header, e.g., • source/destination IP address (32 bits) • source/destination port number (16 bits) • TOS byte (8 bits) • Type of protocol (8 bits) • In general fields are specified by range istoica@cs.berkeley.edu

Example of Classification Rules • Access-control in firewalls • Deny all e-mail traffic from ISP-X to Y • Policy-based routing • Route IP telephony traffic from X to Y via ATM • Differentiate quality of service • Ensure that no more than 50 Mbps are injected from ISP-X istoica@cs.berkeley.edu

Characteristics of Real Classifiers • Results are collected over 793 packet classifiers from 101 ISPs, with a total of 41,505 rules • Classifiers do not contain many rules: mean = 50 rules, max = 1734 rules, only 0.7% contain over 1000 rules • Many fields are specified by range, e.g., greater than 1023, or 20-24 • 14% of classifiers had a rule with a non-contiguous mask ! • Rules in the same classifier tend to share the same fields • 8% of the rules are redundant, i.e., they can be eliminated without changing classifier’s behavior istoica@cs.berkeley.edu

Example • Two-dimension space (i.e., classification based on two fields) • Complexity depends of the layout (i.e., how many distinct regions are created) istoica@cs.berkeley.edu

Hard Problem • Even if regions don’t overlap, with n rules and F fields we have the following lower-bounds • O(log n) time and O(nF) space • O(log F-1 n) time and O(n) space istoica@cs.berkeley.edu

Simplifying Assumptions • In practice, you get the average not the worst-case, e.g., number of overlapping regions for the largest classifier 4316 vs. theoretical worst case 10 13 • The number of rules is reasonable small, i.e., at most several thousands • The rules do not change often istoica@cs.berkeley.edu

Recursive Flow Classification (RFC) Algorithm • Problem formulation: • Map S bits (i.e., the bits of all the F fields) to T bits (i.e., the class identifier) • Main idea: • Create a 2S table with pre-computed values; each entry would contain the class identifier • Only one memory access needed • …but this is impractical  require huge memory istoica@cs.berkeley.edu

RFC Algorithm • Use recursion: trade speed (number of memory accesses) for memory footprint istoica@cs.berkeley.edu

The RFC Algorithm • Split the F fields in chuncks • Use the value of each chunck to index into a table • Indexing is done in parallel • Combine results from previous phase, and repeat • In the final phase we obtain only one value istoica@cs.berkeley.edu

Example of Packet Flow in RFC istoica@cs.berkeley.edu

Complete Example • Four fields  six chunks • Source and destination IP addresses  two chuncks each • Protocol number  one chunck • Destination port number  one chunck istoica@cs.berkeley.edu

Complete Example indx=c10*5+c11 indx=c02*6+c03*3+c05 istoica@cs.berkeley.edu

indx=c10*5+c11 istoica@cs.berkeley.edu

RFC Lookup Performance • Dataset: classifiers used in practice • Hardware: 31.25 millions pps using three stage pipeline, and 4-bank 64 Mb SRAMs at 125 MHz • Software: > 1million pps on a 333 MHz Pentium istoica@cs.berkeley.edu

RFC Scalling • RFC does not handle well large (general) classifiers • As the number of rules increases, the memory requirements increase dramatically, e.g., for 1500 rules you may need over 4.5 MB with a three stage classifier • Proposed solution: adjacency groups • Idea: group rules that generate the same actions and use same fields • Problems: can’t tell which rule was matched istoica@cs.berkeley.edu

Summary • Routing lookup and packet classification  two of the most important challenges in designing high speed routers • Very efficient algorithms for routing lookup  possible to do lookup at the line speed • Packet classification still an area of active research • Key difficulties in designing packet classification: • Requires multi-field classification which is an inherently hard problem • If we want per flow QoS insertion/deletion need also to be fast • Harder to make update-lookup tradeoffs like Lulea’s algorithm istoica@cs.berkeley.edu

RFC Algorithm: Example • Phase 0: • Possible values for destination port number: 80, 20-21, >1023, * • Use two bits to encode • Reduction: 162 • Possible values for protocol: udp, tcp, * • Use two bits to encode • Reduction: 82 • Phase 1: • Concatenate from phase 1, five possible values: {80,udp}, {20-21,udp}, {80,tcp}, {>1023,tcp}, everything else • Use three bits to encode • Reduction 43 istoica@cs.berkeley.edu

CS 268: Lectures 13/14 (Route Lookup and Packet Classification)